ENGINEERING MATHEMATICS 4 (BDA 34003) Lecture Module 7: Eigenvalue and Eigenvector Waluyo Adi Siswanto Universiti Tun Hussein Onn Malaysia This work is licensed under a Creative Commons Attribution 3.0 License.
Topics Eigenvalue and eigenvector in Engineering Characteristics equation Power Method Shifted Power Method Inverse Power Method Lecture Module 7 BDA34003 2
Learning Outcomes Students have the knowlegde of eigenvalue and eigenvector in engineering applications Students will be able to determine natural frequencies of structural engineering problems Students will be able to decide the appropriate method for a particular structural engineering problem Lecture Module 7 BDA34003 3
Eigenvalue and Eigenvector in Engineering The bridge was shaken by the wind. This external shaking frequency coincided or very close to the natural frequency of the bridge. Since the external frequency met with the natural frequency, the displacement amplitudes were then amplified (resonance). The bridge was collapse because of this resonance. Lecture Module 7 BDA34003 4
Eigenvalue and Eigenvector Magnification of amplitude at resonance Vibration on its Natural frequency External excitation Resonance result Therefore you have to know the natural frequency, so you know dangerous external citations that will damage the structure because of resonance phenomenon Lecture Module 7 BDA34003 5
Eigenvalue and Eigenvector Natural Frequencies = characteristics Every system has its own natural frequencies. Therefore natural frequencies shows the characteristics (German = eigen) of the system. The equation representing the characteristics is then called characteristics equation Every system has its own characteristics f n =0 Hz mode = f n =0 Hz mode={,} T f n2 =25 Hz mode 2={, 0.8} T Lecture Module 7 BDA34003 6
Eigenvalue and Eigenvector Natural Frequencies and modeshapes Natural frequencies are obtained from the eigenvalues Mode shapes are obtained from the eigenvectors Eigenvalue no Eigenvector no f n =0 Hz mode={,} T Eigenvalue no 2 f n2 =25 Hz Eigenvector no 2 mode 2={, 0.8} T Lecture Module 7 BDA34003 7
Eigenvalue and Eigenvector Characteristics equation From this simple vibration illustration, we will see the characteristics equation m x + (k + k 2 ) x k 2 x 2 =0 m 2 x 2 + k 2 ( x 2 x )=0 m k k 2 x x + (k + k 2 ) x m k 2 x m 2 =0 x 2 + k 2 (x m 2 x )=0 2 m 2 x 2 [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 2} {x x + [ 0 ]{ 0 x x 2} =0 Lecture Module 7 BDA34003 8
Characteristics equation m k x [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 2} {x x + [ 0 ]{ 0 x x 2} =0 sin representing x =V sin(ω t) x 2 =V 2 sin(ω t) fluctuation x = V ω 2 sin(ωt) x 2 = V 2 ω 2 sin(ωt) m 2 k 2 x 2 [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 2} {V V + [ ω2 0 2]{ V 2} 0 ω V =0 Lecture Module 7 BDA34003 9
Characteristics equation m k k 2 x [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 2} {V V + [ ω2 0 2]{ V 2} 0 ω V =0 2} {V V [ ω2 0 λ 0 ω 2]{ V V 2} =0 m 2 x 2 _ [ A] {V } λ[ I ] {V } = 0 Standard eigenvalue-eigenvector problem Lecture Module 7 BDA34003 0
Characteristics equation _ [ A] {V } λ[ I ] {V } = 0 [[ A] λ [ I ]]{V }=0 Since {V } 0 [ A] λ [ I ] =0 In mathematics is called Characteristics equation λ {V } eigenvalue eigenvector Lecture Module 7 BDA34003
Example 7- k =20 k =0 m = m 2 = m k k 2 x By using characteristics equation: Find the eigenvalues of the system Find the natural (resonance) frequencies of the system Find the eigenvectors of the system Draw the mode shape corresponding to to each natural frequency m 2 x 2 Lecture Module 7 BDA34003 2
k =20 k 2 =0 m = m 2 = and λ=ω 2 k m k 2 m 2 x x 2 [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 [ 30 2} {V V [ ω2 0 2]{ V 2} 0 ω V =0 0 0 0 ]{ V V 2} [ λ 0 0 λ]{ V V 2} =0 [(30 λ) 0 0 (0 λ)]{ V 2} V =0 (30 λ) 0 0 (0 λ) =0 Characteristics equation Lecture Module 7 BDA34003 3
The characteristics equation can be expanded: (30 λ) 0 0 (0 λ) =0 m k x (30 λ)(0 λ) ( 0)( 0)=0 300 0 λ 30 λ+ λ 2 00=0 λ 2 40 λ+ 200=0 k 2 λ =5.8579 λ 2 =34.42 Eigenvalue Eigenvalue 2 m 2 x 2 In SMath you can write (to solve a polynomial): Lecture Module 7 BDA34003 4
λ =5.8579 λ 2 =34.42 Eigenvalue Eigenvalue 2 m k x Angular natural frequency: ω =2.4203 rad/s ω 2 =5.843 rad/s k 2 Natural frequency: f =0.3852 Hz m 2 x 2 f 2 =0.93 Hz Lecture Module 7 BDA34003 5
λ =5.8579 Eigenvalue k m k 2 m 2 x x 2 [(30 λ) 0 0 (0 λ)]{ V 2} V =0 Normalizing V = (30 λ )V =0V 2 V 2 =2.442 Eigenvector { V V 2} = { 2.442} Lecture Module 7 BDA34003 6
λ 2 =34.42 Eigenvalue 2 k m k 2 m 2 x x 2 [(30 λ) 0 0 (0 λ)]{ V 2} V =0 Normalizing V = (30 λ )V =0V 2 V 2 = 0.442 Eigenvector 2 { V V 2} = { 0.442} Lecture Module 7 BDA34003 7
Physical interpretation of eigenvectors: Eigenvector { V V 2} = { 2.442} Eigenvector 2 { V V 2} = { 0.442} k m x k 2 m 2 x 2 2.442 0.442 Lecture Module 7 BDA34003 8
Example 7-2 k =20 k =0 m = m 2 = m k k 2 x Use FreeMat Find the eigenvalues of the system Find the natural (resonance) frequencies of the system Find the eigenvectors of the system Draw the mode shape corresponding to to each natural frequency m 2 x 2 Lecture Module 7 BDA34003 9
λ =5.8579 λ 2 =34.42 Eigenvalue Eigenvalue 2 Angular natural frequency: ω =2.4203 ω 2 =5.843 rad/s rad/s Natural frequency: f =0.3852 f 2 =0.93 Hz Hz Lecture Module 7 BDA34003 20
λ =5.8579 Eigenvalue { V V 2} = { 0.3827 0.9239} { V V 2} = { 2.442} 2.442 Lecture Module 7 BDA34003 2
λ =34.42 Eigenvalue 2 { V V 2} = { 0.9239 0.3827 } { V V 2} = { 0.442} 0.442 Lecture Module 7 BDA34003 22
Example 7-3 x x 2 x 3 k k k m m m Develop the characteristics equation of the system above Find the eigenvalues of the system Find the natural (resonance) frequencies of the system Find the eigenvectors of the system Draw the mode shape corresponding to to each natural frequency (Use FreeMat to verify the eigenvalues and eigenvectors) Lecture Module 7 BDA34003 23
x x 2 x 3 k k k m m m [ 2k k 0 k 2k k x ]{x 3}+[ m 0 0 2 0 m 0 ẍ 2 0 k k x 0 0 m]{ẍ ẍ 3}=0 x =V sin(ω t) x 2 =V 2 sin(ω t) x 3 =V 3 sin(ω t) [ 2k 2 k 0 ω 0 0 k 2k k V 0 k k ]{V 2 0 m ω V 3} [m 2]{V 2 0 V 2 0 0 m ω V 3}=0 Lecture Module 7 BDA34003 24
[ 2 0 2 V 0 ]{V 3} [ 2 mω 0 0 k mω 2 2 0 0 k V m ω 2 0 0 k ]{V V 2 V 3}=0 λ= mω2 k [ 2 λ 0 2 λ V 2 0 λ]{v V 3}=0 The characteristics equation 2 λ 0 2 λ 0 λ =0 Lecture Module 7 BDA34003 25
2 λ 0 2 λ 0 λ =0 (2 λ)[(2 λ)( λ) ( )( )] ( )[( )( λ) 0]=0 λ 3 + 5 λ 2 6 λ+ =0 λ =0.98 Eigenvalue λ 2 =.555 Eigenvalue 2 λ 2 =3.247 Eigenvalue 3 Lecture Module 7 BDA34003 26
λ =0.98 Eigenvalue λ 2 =.555 Eigenvalue 2 λ 2 =3.247 Eigenvalue 3 natural frequency λ= mω2 k ω =0.98 k / m ω 2 =.555 k / m ω 3 =3.247 k / m rad/s rad/s rad/s f = 0.98 2 π k / m f 2 =.555 2 π k / m Hz Hz f 3 = 3.247 2 π k / m Hz Lecture Module 7 BDA34003 27
λ =0.98 Eigenvalue [ 2 λ 0 2 λ V 2 normalized V = 0 λ]{v V 3}=0 (2 λ)v 2 V 3 =0 ( )V V 2 + ( λ)v 3 =0.809V 2 V 3 = V 2 + 0.809V 3 =0 V 2 =.8023 V 3 =2.2475 Eigenvector {V V 2 V 3}= {.8023 2.2475} Lecture Module 7 BDA34003 28
λ 2 =.555 Eigenvalue 2 [ 2 λ 0 2 λ V 2 normalized V = 0 λ]{v V 3}=0 (2 λ)v 2 V 3 =0 ( )V V 2 + ( λ)v 3 =0 0.445V 2 V 3 = V 2 0.555V 3 =0 V 2 =0.445 V 3 = 0.809 {V V 3}={ Eigenvector 2 2 0.445 V 0.809} Lecture Module 7 BDA34003 29
λ 3 =3.247 Eigenvalue 3 [ 2 λ 0 2 λ V 2 normalized V = 0 λ]{v V 3}=0 (2 λ)v 2 V 3 =0 ( )V V 2 + ( λ)v 3 =0.247V 2 V 3 = V 2 2.247V 3 =0 V 2 =.2469 V 3 =0.5549 {V V 3}={ } Eigenvector 3 2.2469 V 0.5549 Lecture Module 7 BDA34003 30
x x 2 x 3 k k k m m m Mode Mode 2 {V V 2 V 3}= {.8023 2.2475} {V V 3}={ 2 V 0.445 0.809} {V V 2 V 3}= { } Mode 3.2469 0.5549 Lecture Module 7 BDA34003 3
If you normalise the eigenvectors: Eigenvector Eigenvector 2 Eigenvector 3 {V V 2 V 3}= {.8023 2.2475} {V V 3}={ 2 V 0.445 0.809} {V V 2 V 3}= { }.2469 0.5549 Lecture Module 7 BDA34003 32
This equation _ [ A] {V } λ[ I ] {V } = 0 Power Method can be rewritten [ A]{V }=λ {V } So that the eigenvalue can be calculated from the eigenvector In iterative manner {V } k+ [ A]{V }k = λ k+ This method will find the highest eigenvalue and its corresponding eigenvector This iteration requires an initial trial eigenvector {V } k In Engineering, this method is called Matrix Iteration Method Lecture Module 7 BDA34003 33
Example 7-4 Find the eigenvalue and the corresponding eigenvector in Example 7-3 by implementing Power Method (Matrix Iteration Method) Use the initial eigenvector {V }=( ) T Stop the iteration when λ k+ λ k < 0.008 Lecture Module 7 BDA34003 34
The system has been identified: [ 2 0 2 V 0 ]{V 3} [ 2 mω 0 0 k mω 2 2 0 0 k V m ω 2 0 0 k ]{V V 2 V 3}=0 Therefore [ A]=[ 2 0 2 0 ] Lecture Module 7 BDA34003 35
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Continue. until convergence satisfied Lecture Module 7 BDA34003 37
This equation Shifted Power Method _ [ A] {V } λ[ I ] {V } = 0 [ A]{V }=λ {V } [ A]{V } s[ I ]{V }=λ{v } s {V } can be shifted with a scalar value s [ A s[ I ]]{V }=(λ s){v } [ A s ]{V }=λ s {V } So that the eigenvalue can be calculated from the eigenvector In iterative manner {V } k+ = [ A s]{v } k λ s k+ λ=λ s + s This method will find the lowest eigenvalue and its corresponding eigenvector In some vibration books [ A s ] is called the deflated matrix Lecture Module 7 BDA34003 38
Example 7-5 Continue Example 7-4 to find the lowest eigenvalue Lecture Module 7 BDA34003 39
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Continue. until convergence satisfied Lecture Module 7 BDA34003 4
This equation Inverse Power Method _ [ A] {V } λ[ I ] {V } = 0 [ A]{V }=λ {V } can be modified by multiplying [ A] [ A][ A] {V }=λ [ A] {V } {V }=λ[a] {V } So that the eigenvalue can be calculated from the eigenvector In iterative manner {V } k+ = [ A] {V } k λ k+ This method will find the lowest eigenvalue and its corresponding eigenvector λ= λ k+ This iteration requires an initial trial eigenvector {V } k Standard Matrix Iteration Method In Engineering, this method is called Standard Because it gives the lowest eigenvalue, which is the standard information in vibration. Lecture Module 7 BDA34003 42
Example 7-6 Do similar problem Example 7-4 by using the Standard Matrix Iteration (Inverse Power Method) Lecture Module 7 BDA34003 43
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Continue. until convergence satisfied Lecture Module 7 BDA34003 45
Student Activity Lecture Module 7 BDA34003 46
Problem 7-A x x 2 x 3 k k k m m m k=2 m=2 Derive the dynamics equation of motion of the system Find the natural frequencies and the corresponding mode shapes by using - characteristics equation - starndard iteration method - verify your result using freemat Lecture Module 7 BDA34003 47
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